## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.8 - Inequalities - 1.8 Exercises - Page 89: 57

#### Answer

The solution is $(-2,0)\cup(2,\infty)$ The graph is

#### Work Step by Step

$x^{3}-4x\gt0$ Factor the left side: $x(x^{2}-4)\gt0$ $x(x-2)(x+2)\gt0$ Find the intervals. The factors are $x$, $x-2$ and $x+2$. Set them equal to $0$ and solve for $x$: $x=0$ $x-2=0$ $x=2$ $x+2=0$ $x=-2$ The factors are zero when $x=0,2,-2$. These three numbers divide the real line into the following intervals: $(-\infty,-2)$ $,$ $(-2,0)$ $,$ $(0,2)$ $,$ $(2,\infty)$ Elaborate a diagram, using test points to determine the sign of each factor in each interval: (refer to the attached image below) It can be seen from the diagram that the inequality is satisfied on the intervals $(-2,0)$ and $(2,\infty)$. Also, the inequality involves $\gt$ so the endpoints do not satisfy the inequality. The solution is $(-2,0)\cup(2,\infty)$

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